Minimum Area of a Continuous Thickener: 23-Chem-A2 Unit Operations and Separation Processes, December 2018, Q1

Question A1: Minimum Area of a Continuous Thickener (25 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Concentration $X$ (kg water/kg solids)	Sedimentation rate $u_c$ (mm/s)
5.0 (feed)	0.20
4.2	0.12
3.7	0.094
3.1	0.07
2.5	0.05
Underflow $X_u$	1.5
Solids feed rate $W$	4788 kg/hr $=1.33$ kg/s

Find. The minimum settling area $A$ (m²) that lets the continuous thickener discharge underflow at $X_u=1.5$ while clarifying the overflow.

Figure A1 — In each horizontal layer the water released by thickening must rise faster than the settling solids. The required area is the largest of the layer-by-layer demands (Coe–Clevenger); the controlling layer sets the minimum thickener area.

Approach. By the Coe–Clevenger method the area needed to clarify a layer at concentration $X$ is $A=\dfrac{W\,(X-X_u)}{\rho_w\,u_c}$; evaluate it for every test concentration and take the maximum, because the layer with the greatest area demand controls.

Solids mass rate in SI. $W=\dfrac{4788}{3600}=\boxed{1.33\ \text{kg solids/s}}$.
Coe–Clevenger area per layer. The upward volumetric flow of released water in a layer is $W(X-X_u)/\rho_w$; dividing by the settling velocity gives the area, $$A=\frac{W\,(X-X_u)}{\rho_w\,u_c},\qquad \rho_w=1000\ \text{kg/m}^3.$$ Applying it to each test point (with $u_c$ converted mm/s → m/s):

$X$ (kg w/kg s)	$X-X_u$	$u_c$ (m/s)	$A$ (m²)
5.0	3.5	$2.0\times10^{-4}$	23.27
4.2	2.7	$1.2\times10^{-4}$	29.93
3.7	2.2	$\mathbf{9.4\times10^{-5}}$	31.13 ← max
3.1	1.6	$7.0\times10^{-5}$	30.40
2.5	1.0	$5.0\times10^{-5}$	26.60

Controlling layer and minimum area. The largest demand occurs at $X=3.7$, so $$A_\text{min}=\frac{1.33\,(3.7-1.5)}{1000\,(9.4\times10^{-5})}=\boxed{31.1\ \text{m}^2}.$$ A thickener smaller than this would let solids escape in the overflow at the $X=3.7$ layer; a larger unit is safe but wasteful. The corresponding diameter is $D=\sqrt{4A/\pi}=6.3$ m.

Quantity	Value
Solids mass rate $W$	1.33 kg/s
Controlling concentration	$X = 3.7$ kg water/kg solids
Minimum thickener area	31.1 m² ($D \approx 6.3$ m)

Concept & Theory

Topic. Sizing a continuous gravity thickener by the Coe–Clevenger (batch-flux) method.

Key relations. Layer area $A=W(X-X_u)/(\rho_w u_c)$; controlling area $A_\text{min}=\max_X A(X)$; the equivalent Kynch/Yoshioka construction locates the same limiting flux graphically.

Why this works. In steady operation the water leaving with the underflow is fixed by $X_u$; every intermediate layer must pass upward the excess water $(X-X_u)$ carried in with the solids. That upward liquid velocity cannot exceed the local hindered-settling velocity $u_c$, or solids would be swept over the weir. Each concentration therefore imposes an area requirement, and the thickener must satisfy the most demanding one — hence the maximum, not the average.

Common pitfall. Using only the feed concentration (here $X=5.0$ gives just 23.3 m², a 25% undersize) instead of scanning all layers; or forgetting the mm/s → m/s conversion, which mis-scales the area by $10^3$.

Source. Coulson & Richardson Vol. 2 ch. 5 (sedimentation, Coe–Clevenger); McCabe ch. 29; Perry's §18 (thickeners & clarifiers).

Question 1 of 6: Minimum Area of a Continuous Thickener

Question A1: Minimum Area of a Continuous Thickener (25 points)